# Integration by Parts

## Related calculator: Integral (Antiderivative) Calculator with Steps

It is easy to compute the integral $\int{{e}}^{{x}}{d}{x}$, but how to handle integrals like $\int{x}{{e}}^{{x}}{d}{x}$?
In general, if you have under the integral sign a product of functions that can be easily integrated separately, you should use integration by parts.

Formula for integration by parts: $\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$.

Proof

Using the product rule, we have that ${\left({f{{\left({x}\right)}}}{g{{\left({x}\right)}}}\right)}'={f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}+{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}$.

Integrating both sides gives: $\int{\left({f{{\left({x}\right)}}}{g{{\left({x}\right)}}}\right)}'{d}{x}=\int{\left({f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}+{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}\right)}{d}{x}$.

This can be rewritten as ${f{{\left({x}\right)}}}{g{{\left({x}\right)}}}=\int{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}{d}{x}+\int{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}{d}{x}$ or $\int{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}{d}{x}={f{{\left({x}\right)}}}{g{{\left({x}\right)}}}-\int{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}{d}{x}$.

If we take ${u}={f{{\left({x}\right)}}}$ and ${v}={g{{\left({x}\right)}}}$, we have that ${d}{u}={f{'}}{\left({x}\right)}{d}{x}$ and ${d}{v}={g{'}}{\left({x}\right)}{d}{x}$, and the above formula can be rewritten using the substitution rule as $\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$.

As can be seen, integration by parts corresponds to the product rule (just like the substitution rule corresponds to the chain rule).

In fact, every differentiation rule has a corresponding integration rule, because these processes are the inverse of each other.

Example 1. Evaluate $\int{x}{{e}}^{{x}}{d}{x}$.

Let ${u}={x}$ and ${d}{v}={{e}}^{{x}}{d}{x}$. Then, ${d}{u}={\left({x}\right)}'{d}{x}={d}{x}$, and ${v}=\int{{e}}^{{x}}{d}{x}={{e}}^{{x}}$.

So, $\int{x}{{e}}^{{x}}{d}{x}=\int\overbrace{{\left({x}\right)}}^{{u}}\overbrace{{\left({{e}}^{{x}}{d}{x}\right)}}^{{{d}{v}}}=\overbrace{{\left({x}\right)}}^{{u}}\overbrace{{\left({{e}}^{{x}}\right)}}^{{v}}-\int\overbrace{{\left({{e}}^{{x}}\right)}}^{{v}}\overbrace{{\left({d}{x}\right)}}^{{{d}{u}}}={x}{{e}}^{{x}}-{{e}}^{{x}}+{C}$.

Note that it is very important to choose appropriate ${u}$ and ${v}$, because a wrong choice will only complicate the integral.

For example, assume that, instead of choosing ${u}={x}$ and ${d}{v}={{e}}^{{x}}{d}{x}$ in the above example, we choose ${u}={{e}}^{{x}}$ and ${v}={x}{d}{x}$. Then, ${d}{u}={\left({{e}}^{{x}}\right)}'{d}{x}={{e}}^{{x}}{d}{x}$, and ${v}=\int{x}{d}{x}=\frac{{{x}}^{{2}}}{{2}}$.

So, $\int{x}{{e}}^{{x}}{d}{x}=\int\overbrace{{\left({{e}}^{{x}}\right)}}^{{u}}\overbrace{{\left({x}{d}{x}\right)}}^{{{d}{v}}}=\overbrace{{\left({{e}}^{{x}}\right)}}^{{u}}\overbrace{{\left(\frac{{1}}{{2}}{{x}}^{{2}}\right)}}^{{v}}-\int\overbrace{{\left(\frac{{1}}{{2}}{{x}}^{{2}}\right)}}^{{v}}\overbrace{{\left({{e}}^{{x}}{d}{x}\right)}}^{{{d}{u}}}$.

Although this equation is true, the integral $\int\frac{{1}}{{2}}{{x}}^{{2}}{{e}}^{{x}}{d}{x}$ is more difficult to evaluate than the integral we started with.

Example 2. Evaluate $\int{{x}}^{{2}}{\cos{{\left({x}\right)}}}{d}{x}$.

Notice that ${{x}}^{{2}}$ becomes simpler when differentiated. Therefore, let ${u}={{x}}^{{2}}$ and ${d}{v}={\cos{{\left({x}\right)}}}{d}{x}$. Then, ${d}{u}={\left({{x}}^{{2}}\right)}'{d}{x}={2}{x}{d}{x}$, and ${v}=\int{\cos{{\left({x}\right)}}}{d}{x}={\sin{{\left({x}\right)}}}$.

So, $\int{{x}}^{{2}}{\cos{{\left({x}\right)}}}{d}{x}={{x}}^{{2}}{\sin{{\left({x}\right)}}}-\int{2}{x}{\sin{{\left({x}\right)}}}{d}{x}={{x}}^{{2}}{\sin{{\left({x}\right)}}}-{2}\int{x}{\sin{{\left({x}\right)}}}{d}{x}$.

We've obtained a simpler integral, but still it is not obvious. Therefore, we apply integration by parts once more to the integral $\int{x}{\sin{{\left({x}\right)}}}{d}{x}$:

let ${u}={x}$ and ${d}{v}={\sin{{\left({x}\right)}}}{d}{x}$; then, ${d}{u}={\left({x}\right)}'{d}{x}={d}{x}$, and ${v}=\int{\sin{{\left({x}\right)}}}{d}{x}=-{\cos{{\left({x}\right)}}}$.

So, $\int\overbrace{{\left({x}\right)}}^{{u}}\overbrace{{\left({\sin{{\left({x}\right)}}}{d}{x}\right)}}^{{{d}{v}}}=\overbrace{{\left({x}\right)}}^{{u}}\overbrace{{\left(-{\cos{{\left({x}\right)}}}\right)}}^{{v}}-\int\overbrace{{\left(-{\cos{{\left({x}\right)}}}\right)}}^{{{v}}}\overbrace{{\left({d}{x}\right)}}^{{{d}{u}}}=-{x}{\cos{{\left({x}\right)}}}+\int{\cos{{\left({x}\right)}}}{d}{x}=$

$=-{x}{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}}+{C}$.

Finally, $\int{{x}}^{{2}}{\cos{{\left({x}\right)}}}{d}{x}={{x}}^{{2}}{\sin{{\left({x}\right)}}}-{2}\int{x}{\sin{{\left({x}\right)}}}{d}{x}={{x}}^{{2}}{\sin{{\left({x}\right)}}}-{2}{\left(-{x}{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}}+{C}\right)}=$

$={{x}}^{{2}}{\sin{{\left({x}\right)}}}+{2}{x}{\cos{{\left({x}\right)}}}-{2}{\sin{{\left({x}\right)}}}+{C}_{{1}}$ where ${C}_{{1}}=-{2}{C}$.

Example 2 shows that in some cases we need to apply integration by parts more than once.

Example 3. Evaluate $\int{\ln{{\left({x}\right)}}}{d}{x}$.

Here, there is only one choice for ${u}$ and ${v}$, namely ${u}={\ln{{\left({x}\right)}}}$ and ${d}{v}={d}{x}$; so,${d}{u}={\left({\ln{{\left({x}\right)}}}\right)}'{d}{x}=\frac{{1}}{{x}}{d}{x}$, and ${v}=\int{d}{x}={x}$.

Therefore, $\int\overbrace{{\left({\ln{{\left({x}\right)}}}\right)}}^{{u}}\overbrace{{\left({d}{x}\right)}}^{{{d}{v}}}=\overbrace{{\left({\ln{{\left({x}\right)}}}\right)}}^{{u}}\overbrace{{\left({x}\right)}}^{{v}}-\int\overbrace{{\left({x}\right)}}^{{v}}\overbrace{{\left(\frac{{1}}{{x}}{d}{x}\right)}}^{{{d}{u}}}=$

$={x}{\ln{{\left({x}\right)}}}-\int{d}{x}={x}{\ln{{\left({x}\right)}}}-{x}+{C}$.

Example 4. Evaluate $\int{{e}}^{{x}}{\sin{{\left({x}\right)}}}{d}{x}$.

Neither ${{e}}^{{x}}$ nor ${\sin{{\left({x}\right)}}}$ become simpler when differentiated, but we try choosing ${u}={{e}}^{{x}}$ and ${v}={\sin{{\left({x}\right)}}}{d}{x}$.

Then, ${d}{u}={\left({{e}}^{{x}}\right)}'{d}{x}={{e}}^{{x}}{d}{x}$, and ${v}=\int{\sin{{\left({x}\right)}}}{d}{x}=-{\cos{{\left({x}\right)}}}{d}{x}$.

Therefore, $\int{{e}}^{{x}}{\sin{{\left({x}\right)}}}{d}{x}=-{{e}}^{{x}}{\cos{{\left({x}\right)}}}-\int-{\cos{{\left({x}\right)}}}{{e}}^{{x}}{d}{x}=-{{e}}^{{x}}{\cos{{\left({x}\right)}}}+\int{{e}}^{{x}}{\cos{{\left({x}\right)}}}{d}{x}$.

The integral obtained is not simpler than the original one, but it is not more difficult either. So, we apply integration by parts once more: let ${u}={{e}}^{{x}}$ and ${d}{v}={\cos{{\left({x}\right)}}}{d}{x}$; then, ${d}{u}={\left({{e}}^{{x}}\right)}'{d}{x}={{e}}^{{x}}{d}{x}$, and ${v}=\int{\cos{{\left({x}\right)}}}{d}{x}={\sin{{\left({x}\right)}}}$.

So, $\int{{e}}^{{x}}{\cos{{\left({x}\right)}}}{d}{x}={{e}}^{{x}}{\sin{{\left({x}\right)}}}-\int{{e}}^{{x}}{\sin{{\left({x}\right)}}}{d}{x}$.

We've obtained the initial integral! It seems that we obtained nothing because we've arrived at $\int{{e}}^{{x}}{\sin{{\left({x}\right)}}}{d}{x}$, which is where we had started.

However, since $\int{{e}}^{{x}}{\sin{{\left({x}\right)}}}{d}{x}=-{{e}}^{{x}}{\cos{{\left({x}\right)}}}+\int{{e}}^{{x}}{\cos{{\left({x}\right)}}}{d}{x}$, it can be stated that

$\int{{e}}^{{x}}{\sin{{\left({x}\right)}}}{d}{x}=-{{e}}^{{x}}{\cos{{\left({x}\right)}}}+{\left({{e}}^{{x}}{\sin{{\left({x}\right)}}}-\int{{e}}^{{x}}{\sin{{\left({x}\right)}}}{d}{x}\right)}$.

This can be regarded as an equation with the unknown variable $\int{{e}}^{{x}}{\sin{{\left({x}\right)}}}{d}{x}$.

This equation can be rewritten as ${2}\int{{e}}^{{x}}{\sin{{\left({x}\right)}}}{d}{x}={{e}}^{{x}}{\left({\sin{{\left({x}\right)}}}-{\cos{{\left({x}\right)}}}\right)}$.

Dividing by 2 and adding the constant of integration yields the final answer: $\int{{e}}^{{x}}{\sin{{\left({x}\right)}}}{d}{x}=\frac{{1}}{{2}}{{e}}^{{x}}{\left({\sin{{\left({x}\right)}}}-{\cos{{\left({x}\right)}}}\right)}+{C}$.

Example 5. Prove the reduction formula.$\int{{\sin}}^{{n}}{\left({x}\right)}{d}{x}=-\frac{{1}}{{n}}{\cos{{\left({x}\right)}}}{{\sin}}^{{{n}-{1}}}{\left({x}\right)}+\frac{{{n}-{1}}}{{n}}\int{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{d}{x}$, where ${n}\ge{2}$ is an integer.

Let ${u}={{\sin}}^{{{n}-{1}}}{\left({x}\right)}$ and ${d}{v}={\sin{{\left({x}\right)}}}{d}{x}$; then,${d}{u}={\left({{\sin}}^{{{n}-{1}}}\right)}'{d}{x}={\left({n}-{1}\right)}{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{\left({\sin{{\left({x}\right)}}}\right)}'{d}{x}={\left({n}-{1}\right)}{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{\cos{{\left({x}\right)}}}{d}{x}$, and ${v}=\int{\sin{{\left({x}\right)}}}{d}{x}=-{\cos{{\left({x}\right)}}}$.

So, $\int{{\sin}}^{{n}}{\left({x}\right)}{d}{x}=\int{{\sin}}^{{{n}-{1}}}{\sin{{\left({x}\right)}}}{d}{x}=-{\cos{{\left({x}\right)}}}{{\sin}}^{{{n}-{1}}}{\left({x}\right)}-\int{\cos{{\left({x}\right)}}}{\left({n}-{1}\right)}{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{\cos{{\left({x}\right)}}}{d}{x}=$

$=-{\cos{{\left({x}\right)}}}{{\sin}}^{{{n}-{1}}}{\left({x}\right)}+{\left({n}-{1}\right)}\int{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{{\cos}}^{{2}}{\left({x}\right)}{d}{x}$.

Since ${{\cos}}^{{2}}{\left({x}\right)}={1}-{{\sin}}^{{2}}{\left({x}\right)}$, it can be stated that${\left({n}-{1}\right)}\int{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{{\cos}}^{{2}}{\left({x}\right)}{d}{x}={\left({n}-{1}\right)}\int{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{\left({1}-{{\sin}}^{{2}}{\left({x}\right)}\right)}{d}{x}=$

$={\left({n}-{1}\right)}\int{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{d}{x}-{\left({n}-{1}\right)}\int{{\sin}}^{{{n}-{2}}}{{\sin}}^{{2}}{\left({x}\right)}{d}{x}=$

$={\left({n}-{1}\right)}\int{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{d}{x}-{\left({n}-{1}\right)}\int{{\sin}}^{{n}}{\left({x}\right)}{d}{x}$.

Therefore,

$\int{{\sin}}^{{n}}{\left({x}\right)}{d}{x}=-{\cos{{\left({x}\right)}}}{{\sin}}^{{{n}-{1}}}{\left({x}\right)}+{\left({n}-{1}\right)}\int{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{d}{x}-{\left({n}-{1}\right)}\int{{\sin}}^{{n}}{\left({x}\right)}{d}{x}$.

This is an equation with an unknown variable $\int{{\sin}}^{{n}}{\left({x}\right)}{d}{x}$.

It can be rewritten as ${n}\int{{\sin}}^{{n}}{\left({x}\right)}{d}{x}=-{\cos{{\left({x}\right)}}}{{\sin}}^{{{n}-{1}}}{\left({x}\right)}+{\left({n}-{1}\right)}\int{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{d}{x}$.

So, finally, we have that $\int{{\sin}}^{{n}}{\left({x}\right)}{d}{x}=-\frac{{1}}{{n}}{\cos{{\left({x}\right)}}}{{\sin}}^{{{n}-{1}}}{\left({x}\right)}+\frac{{{n}-{1}}}{{n}}\int{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{d}{x}$.

The last question in this section is how to calculate definite integrals with the help of integration by parts. In fact, it is very easy: just combine integration by parts with the Newton-Leibniz formula: ${\int_{{a}}^{{b}}}{u}{d}{v}={u}{v}{{\mid}_{{a}}^{{b}}}-{\int_{{a}}^{{b}}}{v}{d}{u}$.

Example 6. Calculate ${\int_{{0}}^{{1}}}{{\tan}}^{{-{1}}}{\left({x}\right)}{d}{x}$.

Let ${u}={{\tan}}^{{-{1}}}{\left({x}\right)}$ and ${d}{v}={d}{x}$; then, ${d}{u}={\left({{\tan}}^{{-{1}}}{\left({x}\right)}\right)}'{d}{x}=\frac{{1}}{{{{x}}^{{2}}+{1}}}{d}{x}$, and ${v}=\int{d}{x}={x}$.

So, ${\int_{{0}}^{{1}}}{{\tan}}^{{-{1}}}{\left({x}\right)}{d}{x}={x}{{\tan}}^{{-{1}}}{\left({x}\right)}{{\mid}_{{0}}^{{1}}}-{\int_{{0}}^{{1}}}\frac{{x}}{{{{x}}^{{2}}+{1}}}{d}{x}=$

$={\left({1}\cdot{{\tan}}^{{-{1}}}{\left({1}\right)}-{0}\cdot{{\tan}}^{{-{1}}}{\left({0}\right)}\right)}-{\int_{{0}}^{{1}}}\frac{{x}}{{{{x}}^{{2}}+{1}}}{d}{x}=\frac{\pi}{{4}}-{\int_{{0}}^{{1}}}\frac{{x}}{{{{x}}^{{2}}+{1}}}{d}{x}$.

To calculate ${\int_{{0}}^{{1}}}\frac{{x}}{{{{x}}^{{2}}+{1}}}{d}{x}$, we use the substitution rule: let ${t}={{x}}^{{2}}+{1}$; then,${d}{t}={\left({{x}}^{{2}}+{1}\right)}'{d}{x}={2}{x}{d}{x}$, or ${x}{d}{x}=\frac{{1}}{{2}}{d}{t}$.

Since ${x}$ is changing from 0 to 1, it can be stated that ${t}$ is changing from ${2}\cdot{0}={0}$ to ${2}\cdot{1}={2}$.

So, ${\int_{{0}}^{{1}}}\frac{{x}}{{{{x}}^{{2}}+{1}}}{d}{x}={\int_{{1}}^{{2}}}\frac{{1}}{{t}}\frac{{1}}{{2}}{d}{t}=\frac{{1}}{{2}}{{\left[{\ln}{\left|{t}\right|}\right]}_{{1}}^{{2}}}=\frac{{1}}{{2}}{\left({\ln{{\left({2}\right)}}}-{\ln{{\left({1}\right)}}}\right)}=$

$=\frac{{1}}{{2}}{\left({\ln{{\left({2}\right)}}}-{0}\right)}=\frac{{1}}{{2}}{\ln{{\left({2}\right)}}}$.

Finally, we can see that ${\int_{{0}}^{{1}}}{{\tan}}^{{-{1}}}{\left({x}\right)}=\frac{\pi}{{4}}-{\int_{{0}}^{{1}}}\frac{{x}}{{{{x}}^{{2}}+{1}}}{d}{x}=\frac{\pi}{{4}}-\frac{{1}}{{2}}{\ln{{\left({2}\right)}}}$.